Initial Value Problems

**Proof_of_life** · September 2nd 2008, 01:48 PM

It's been awhile since I have had calculus and dealing with differential equations and initial value problems. Looking for some help with some easy problems.

Find the solution of the given initial value problem.

(1) $y' + 2y = te^{-2t}$ , $y(1)=0$

(2) $t^3y' + 4t^2y = e^{-t}$ , $y(-1)=0, t < 0$

A walkthrough of these would be greatly appreciated. Thanks in advance

**ThePerfectHacker** · September 2nd 2008, 03:15 PM

Originally Posted by Proof_of_life

(1) $y' + 2y = te^{-2t}$ , $y(1)=0$

The integrating factor is $f(x) = e^{\int 2 dx}= e^{2x}$ .
Therefore the solution is,
$y=\frac{1}{f(x)} \int f(x) xe^{-2x} dx = e^{-2x} \int x dx = \frac{1}{2}x^2e^{-2x} + ke^{-2x}$ .
Now $y(1) = \frac{1}{2}e^{-2} + k e^{-2} = 0 \implies k = -\frac{1}{2}$ .
Thus, $y = \frac{1}{2}x^2e^{-2x} - \frac{1}{2} e^{-2x}$ .

(2) $t^3y' + 4t^2y = e^{-t}$ , $y(-1)=0, t < 0$

A walkthrough of these would be greatly appreciated. Thanks in advance

This one is similar. Just divide out the leading coefficient first.

**Proof_of_life** · September 2nd 2008, 03:34 PM

I was about to reply that I figured out both of these. Thanks for replying though!

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